Yahoo Answers: Answers and Comments for 2 dice tossed What’s the probability the sum is even? The sum is less than 5? The sum is even & less than 5? The sum is even or less than 5? [Mathematics]
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From ladyrose
enCA
Fri, 22 Nov 2019 23:40:03 +0000
3
Yahoo Answers: Answers and Comments for 2 dice tossed What’s the probability the sum is even? The sum is less than 5? The sum is even & less than 5? The sum is even or less than 5? [Mathematics]
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From A.J.: Will show two methods to answer this.
Method #...
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https://ca.answers.yahoo.com/question/index?qid=20191122234003AA65fTu
Sat, 23 Nov 2019 00:25:56 +0000
Will show two methods to answer this.
Method #1 individual analysis
Two dice rolled is equivalent to one die rolled 2x. Each 1,2,3,4,5,6.
Sum is even. First roll
1,3,5 to get even takes 1,3,5 on the second
2,4,6 to get even takes 2,4,6 on the second
Each of the second outcomes is 50%
The combined first is 100%
Total is 50%, that is each of the 6 has a 50% chance of being an even sum on 2 dice.
Many different ways to get to the 50%
Sum is less than 5
1 = 1/6 good is 1,2,3 = 3/6 >> 1/6 x 3/6 = 3/36
2 = 1/6 good is 1,2 = 2/6 >> 1/6 x 2/6 = 2/36
3 = 1/6 good is 1 = 1/6 >> 1/6 x 1/6 = 1/36
4 = 1/6 no good
5 = 1/6 no good
6 = 1/6 no good
(3+2+1)/36 = 6/36 or = 1/6
Even and less than 5
2 and 4 total
1 = 1/6 good is 1,3 = 2/6 >> 1/6 x 2/6 = 2/36
2 = 1/6 good is 2 = 1/6 >> 1/6 x 1/6 = 1/36
3 = 1/6 good is 1 = 1/6 >> 1/6 x 1/6 = 1/36
4 = 1/6 no good
5 = 1/6 no good
6 = 1/6 no good
4/36 = 1/9
The sum is even or less than 5
1 = 1/6 good is 1,2,3,5 = 4/6 >> 1/6 x 4/6 = 4/36
2 = 1/6 good is 1,2,4,6 = 4/6 >> 1/6 x 4/6 = 4/36
3 = 1/6 good is 1,3,5 = 3/6 >> 1/6 x 3/6 = 3/36
4 = 1/6 good is 2,4,6 = 3/6 >> 1/6 x 3/6 = 3/36
5 = 1/6 good is 1,3,5 = 3/6 >> 1/6 x 3/6 = 3/36
6 = 1/6 good is 2,4,6 = 3/6 >> 1/6 x 3/6 = 3/36
4+4+3+3+3+3 = 20/36 = 5/9
Method 2.
Two dice rolled can be 2 through 12 as sum
Two dice rolled is 6x6 =36 equal chance different results
2 = 1,1 only = 1/36
3 = 1,2 and 2,1 = 2/36
4 = 1,3 and 3,1 and 2,2 = 3/36
5 = 1,4 and 4,1 and 2,3 and 3,2 = 4/36
6 = 1,5 and 5,1 and 2,4 and 4,2 and 3,3 = 5/36
7 = 1,6 and 6,1 and 2,5 and 5,2 and 3,4 and 4,3 = 6/36
8 = 2,6 and 6,2 and 3,5 and 5,3 and 4,4 = 5/36
9 = 3,6 and 6,3 and 4,5 and 5,4 = 4/36
10 = 4,6 and 6,4 and 5,5 = 3/36
11 = 5,6 and 6,5 = 2/36
12 = 6,6 = 1/36
Even >> 2,4,6,8,10,12 = 1/36 + 3/36 + 5/36 + 5/36 + 3/36 + 1/36 = 18/36 = 50%
Less than 5 >> 2,3,4 = 1/36 + 2/36 + 3/36 = 6/36 = 1/6
Even and less than 5 >> 2,4 = 1/36 + 3/36 = 4/36 = 1/9
Even or less than 5 >> 2,3,4,6,8,10,12 = 1/36 + 2/36 + 3/36 + 5/36 + 5/36 + 3/36 + 1/36 = 20/36 = 5/9
There are other methods.

From Charles D: There aren't really formulas to calculate ...
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Sat, 23 Nov 2019 00:39:06 +0000
There aren't really formulas to calculate these probabilities. You have to draw the probability distribution and then add the probabilities. This is a very famous problem set. Lots of people have worked this out. In particular at the University of Wisconsin:
Go to the University of Wisconsin web site:
http://pages.stat.wisc.edu/~ifischer/Intro_Stat/Lecture_Notes/4__Classical_Probability_Distributions/4.1__Discrete_Models.pdf
Scroll down until you see the probability distribution for tossing two dice. If you want to understand how to do the problem, read the web page.
To find P(sum is even) add up the probabilities that the number is 2 OR 4 or 6 etc. Looking at the chart, you will see that it is 18/36 = 1/2
To find P( x < 5) add the probabilities that the number is 2 or 3 or 4 =
(1 + 2 + 3)/36 = 6/36 = 1/6
To find P( even and less than 5) add the probabilities that the number is 2 or 4 =( 1 + 3)/36 = 4/36 = 1/9.
To find P( even or less than 5) = P( x < 5 ) = 1/6.
Hope this helps.